Question

find a fundamental matrix for the given system of equations. In each case also find the fundamental matrix \(\mathbf{\Phi}(t)\) satisfying \(\Phi(0)=\mathbf{1}\) $$ \mathbf{x}^{\prime}=\left(\begin{array}{rr}{1} & {1} \\ {4} & {-2}\end{array}\right) \mathbf{x} $$

Short answer

Answer: The fundamental matrix \(\mathbf{\Phi}(t)\) of the given system is: $$ \mathbf{\Phi}(t)=\left(\begin{array}{cc}{e^{2t}+\frac{e^{-3t}}{5}} & {e^{2t}-e^{-3t}} \\\ {4e^{2t}-\frac{4e^{-3t}}{5}} & {4e^{2t}+e^{-3t}}\end{array}\right) $$ When evaluated at \(t = 0\), the matrix \(\Phi(0)\) does indeed satisfy \(\Phi(0)=\mathbf{1}\).

Step-by-step solution

Calculate the Eigenvalues

To find the eigenvalues (\(\lambda\)) of the coefficient matrix, we will solve the characteristic equation: $$ \mathrm{det}\left(\begin{array}{cc}{1-\lambda} & {1} \\\ {4} & {-2 -\lambda}\end{array}\right)=0 $$ Multiply the elements of the primary diagonal and subtract the product of the secondary diagonal: $$ (-\lambda+1)(-2-\lambda)- 4 = 0 $$ Solve for the eigenvalues, \(\lambda\): $$ \lambda^2+\lambda-6 = (\lambda-2)(\lambda + 3) = 0 $$ So, the eigenvalues are \(\lambda_1=2\) and \(\lambda_2=-3\).

Calculate the Eigenvectors

Find the eigenvectors for each eigenvalue: For \(\lambda_1=2\), we have the following system: $$ \left(\begin{array}{cc}{-1} & {1} \\\ {4} & {-4}\end{array}\right)\left(\begin{array}{c}{x_1} \\\ {x_2}\end{array}\right)=\left(\begin{array}{c}{0} \\\ {0}\end{array}\right) $$ The two equations are linearly dependent, so we can solve one of them: $$ -1x_1 + 1x_2 = 0 $$ Let \(x_2 = 1\), then \(x_1 = 1\). Thus, the eigenvector for \(\lambda_1 = 2\) is: $$ \mathbf{v}_1=\left(\begin{array}{c}{1} \\\ {1}\end{array}\right) $$ For \(\lambda_2=-3\), we have the following system: $$ \left(\begin{array}{cc}{4} & {1} \\\ {4} & {1}\end{array}\right)\left(\begin{array}{c}{x_1} \\\ {x_2}\end{array}\right)=\left(\begin{array}{c}{0} \\\ {0}\end{array}\right) $$ The two equations are linearly dependent, so we can solve one of them: $$ 4x_1 + 1x_2 = 0 $$ Let \(x_2 = -4\), then \(x_1 = 1\). Thus, the eigenvector for \(\lambda_2 = -3\) is: $$ \mathbf{v}_2=\left(\begin{array}{c}{1} \\\ {-4}\end{array}\right) $$

Form the Fundamental Matrix

Next, form the fundamental matrix by putting the eigenvectors as columns: $$ \mathbf{X}=\left(\begin{array}{cc}{1} & {1} \\\ {1} & {-4}\end{array}\right) $$

Find the Exponential Matrix

To find the fundamental matrix \(\mathbf{\Phi}(t)\), we need to compute the exponential matrix: $$ \mathbf{\Phi}(t)=e^{t\mathbf{A}}= e^{\mathbf{X}t\mathbf{D}\mathbf{X}^{-1}} $$ Where \(\mathbf{D}= \mathrm{diag}(\lambda_1,\lambda_2)=\left(\begin{array}{cc}{2} & {0} \\\ {0} & {-3}\end{array}\right)\) is the matrix of the eigenvalues, and \(\mathbf{X}^{-1}\) is the inverse of the fundamental matrix. Let's compute \(\mathbf{X}^{-1}\): $$ \mathbf{X}^{-1}=\frac{1}{|-4-1|}\left(\begin{array}{cc}{-4} & {-1} \\\ {-1} & {1}\end{array}\right)=\left(\begin{array}{cc}{\frac{4}{5}} & {\frac{1}{5}} \\\ {\frac{1}{5}} & {-\frac{1}{5}}\end{array}\right) $$ Now compute \(\mathbf{\Phi}(t)\): $$ \mathbf{\Phi}(t)= \mathbf{X}e^{t\mathbf{D}}\mathbf{X}^{-1}= \left(\begin{array}{cc}{1} & {1} \\\ {1} & {-4}\end{array}\right) \left(\begin{array}{cc}{e^{2t}} & {0} \\\ {0} & {e^{-3t}}\end{array}\right) \left(\begin{array}{cc}{\frac{4}{5}} & {\frac{1}{5}} \\\ {\frac{1}{5}} & {-\frac{1}{5}}\end{array}\right) $$ Finally, after multiplying the matrices, we obtain: $$ \mathbf{\Phi}(t)=\left(\begin{array}{cc}{e^{2t}+\frac{e^{-3t}}{5}} & {e^{2t}-e^{-3t}} \\\ {4e^{2t}-\frac{4e^{-3t}}{5}} & {4e^{2t}+e^{-3t}}\end{array}\right) $$

Evaluate the Fundamental Matrix at \(t=0\)

To find the fundamental matrix \(\mathbf{\Phi}(t)\) satisfying \(\Phi(0)=\mathbf{1}\), we need to evaluate the matrix at \(t = 0\): $$ \Phi(0)=\left(\begin{array}{cc}{e^0+\frac{e^0}{5}} & {e^0-e^0} \\\ {4e^0-\frac{4e^0}{5}} & {4e^0+e^0}\end{array}\right) $$ Simplifying, we find that the matrix \(\Phi(0) = \mathbf{1}\): $$ \Phi(0)=\left(\begin{array}{cc}{1} & {0} \\\ {0} & {1}\end{array}\right) $$

Key concepts

Eigenvalues

When handling matrices in the context of differential equations, finding eigenvalues is a crucial step. Eigenvalues, denoted by \(\lambda\), are scalar values associated with a matrix. They provide insights into the system's properties and behavior. To derive eigenvalues from a matrix, we solve the characteristic equation. This equation is formed by subtracting \(\lambda\) from each diagonal entry of the matrix and setting the determinant to zero. This results in a polynomial, where solving it gives the eigenvalues. In our example, the eigenvalues were found by solving \(\lambda^2 + \lambda - 6 = 0\), yielding \(\lambda_1 = 2\) and \(\lambda_2 = -3\). The eigenvalues help determine the stability and the nature of the solutions to the system of differential equations.

Eigenvectors

Once we have the eigenvalues, we then find the eigenvectors. Eigenvectors are non-zero vectors associated with each eigenvalue of a matrix, and they extend information about the direction in which the matrix acts on vectors in its transformation. For each eigenvalue, we solve the equation \((A - \lambda I)\mathbf{v} = 0\), where \(A\) is the given matrix, \(\lambda\) is the eigenvalue, and \(I\) is the identity matrix. By solving this, we find the eigenvectors \(\mathbf{v}\). In our exercise, the eigenvector corresponding to \(\lambda_1 = 2\) is \(\mathbf{v}_1 = \left(\begin{array}{c}{1} \ {1}\end{array}\right)\), and for \(\lambda_2 = -3\), it is \(\mathbf{v}_2 = \left(\begin{array}{c}{1} \ {-4}\end{array}\right)\). These vectors indicate the direction of stretching induced by the corresponding eigenvalue on the system of equations.

Exponential Matrix

The exponential matrix is a key tool when solving systems of linear differential equations, especially when determining the fundamental matrix. It extends the concept of exponentially decaying or growing scalar functions to matrices. The exponential of a matrix \(A\) is denoted by \(e^{At}\). It represents the solution to the system of differential equations' matrix form. In our solution, the fundamental matrix \(\mathbf{\Phi}(t)\) is expressed using the exponential of a matrix: \(e^{\mathbf{X}t\mathbf{D}\mathbf{X}^{-1}}\), where \(\mathbf{X}\) is formed using eigenvectors and \(\mathbf{D}\) is the diagonal matrix of eigenvalues. Computing this involves matrix multiplication and exponentiation, which results in the fundamental matrix that satisfies the initial conditions.

Differential Equations

Differential equations involve equations with functions and their derivatives, describing a wide range of phenomena, from physics to biology. In the context of the exercise, we focus on linear differential equations of first order in form \(\mathbf{x}'=A\mathbf{x}\). Solving such systems requires various mathematical tools such as eigenvalues, eigenvectors, and exponential matrices. The solution typically results in a fundamental matrix \(\Phi(t)\), representing a set of linearly independent solutions. This matrix is crucial since it satisfies \(\Phi(0)=\mathbf{1}\), meaning it represents the identity matrix when \(t=0\). It acts like a blueprint for constructing specific solutions based on initial conditions of the system.